# Interpretation of the phrase “percent reduction relative to”

In the book Freakonomics, the following phrases appear:

"From 1960 to 1985, the number of police officers fell 50 percent relative to the number of crimes.

I interpret the above to mean that $$r_{old} = \dfrac{n_{crimes_{old}}}{n_{police_{old}}}$$ $$r_{new} = \dfrac{n_{crimes_{new}}}{n_{police_{new}}}$$ $$\dfrac{r_{new} - r_{old}}{r_{old}} = 0.5$$

Then, shortly after, this phrase comes up:

By a conservative calculation, this huge expansion of New York's police force would be expected to reduce crime in New York by 18 percent relative to the national average. If you subtract that 18 percent from New York's homicide reduction, thereby discounting the effect of the police-hiring surge, New York no longer leads the nation with its 73.6 percent drop;

I'm having trouble following this. In particular, what does "relative to the national average" mean here? The most natural way for me to interpret it is "relative to the national average crime rate.'

Following that interpretation, and applying the same logic as with the first phrase above, I get the following:

$$r_{old} = \dfrac{r_{crime_{pre-expansion}}} {r_{crime_{avg}}}$$ $$r_{new} = \dfrac{r_{crime_{post-expansion}}}{r_{crime_{avg}}}$$ $$\dfrac{r_{new} - r_{old}}{r_{old}} = 0.82$$

However, it does not seem right then to discuss simply lopping off 18 percent from the New York homicide rate, unless the NY homicide rate is also implicitly "relative to the national average homicide rate."

However, expanding out the previous equations, we get: $$\dfrac{r_{crime_{post-expansion}} - r_{crime_{pre-expansion}}} {r_{crime_{pre-expansion}}} = 0.82$$

Since this relationship holds, subtracting 18 percent from New York's homicide numbers seems more acceptable. But, this makes the phrase "relative to the national average" meaningless, since it just gets cancelled out, right?

How should I mathematically interpret the phrase, "relative to the national average" here?